;实验15
assume cs:code
stack segment
db 128 dup (0)
stack ends
code segment
start:
mov ax,stack
mov ss,ax
mov sp,128
push cs
pop ds
mov ax,0
mov es,ax
mov si,offset int9 ;安装新的int9
mov di,204h
mov cx,offset int9end - offset int9
cld
rep movsb
push es:[9*4] ;保存原int9
pop es:[200h]
push es:[9*4+2]
pop es:[202h]
cli
mov WORD ptr es:[9*4],204h ;将新的int9送入中断向量表
mov WORD ptr es:[9*4+2],0
sti
;为了看到效果,加入以下代码
;begin
mov dx,0ffffh
mov ax,0ffffh
kkk:
sub ax,1
sbb dx,0
cmp dx,0
jne kkk
cmp ax,0
jne kkk
mov ah,1
int 21h
mov ax,4c00h
int 21h
;end
int9:
push ax
push bx
push cx
push es
in al,60h
pushf
call DWORD ptr cs:[200h]
cmp al,9eh ;判断al是否为A的断码,既松开按键所产生的扫描码
jne int9ret
mov ax,0b800h
mov es,ax
mov bx,2
mov cx,2000
s:
mov BYTE ptr es:[bx],'A'
add bx,2
loop s
int9ret:
pop es
pop cx
pop bx
pop ax
iret
int9end:
nop
code ends
end start
:P